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uysha [10]
3 years ago
8

When spray is applied to a car, the paint has a negative charge and the surface of the car has a positive charge. Some processes

use a negatively charged paint and a grounded object. Explain why this also works.
Physics
1 answer:
RUDIKE [14]3 years ago
4 0

Answer:

Based on the properties electrically charged particles, we have that unlike charges attract and like charges repel each other. In order for proper application of an electrostatically negatively charged paint to be properly applied on the metal body surface of a vehicle, require that for attraction, the surface of the vehicle should be grounded and positively charged so as to effectively attract the negatively charged paint particles as it exits the nozzle, to form a strong attachment with the positively charged surface of the vehicle

Explanation:

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Answer:

t = 106π / 30*2.1

Explanation:

w_{i} = 1.06*10^{2}    => 106

    => 106 x 2π/60

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∝ = -2.1 rad/sec²

w_{f} => 0

w_{f} = w_{i} + ∝t

∴ (w_{f} - w_{i}) / ∝ = t

t = 106π / 30*2.1

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3 years ago
Superman is flying 54.5 m/s when he sees
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348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations d=v_{0} t+\frac{at^{2} }{2} and v_{f} =v_{0} +at where d is distance, v_{0} is the initial velocity, v_{f} is the final velocity, t is time, and a is aceleration.

Superman's initial velocity is v_{0}=54.5\frac{m}{s}, and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know d, v_{0} and t, we have to find the aceleration a in order to find v_{f}.

From the equation d=v_{0} t+\frac{at^{2} }{2} we have to clear a, getting the equation as follows: a=\frac{2(d-v_{0}t) }{t^{2} }.

Substituting the values:

a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}

To find v_{f} we use the equation v_{f} =v_{0} +at.

Substituting the values:

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https://physicsabout.com/acceleration-and-velcoity/

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Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ,
makkiz [27]

Answer: \theta=cos^{-1}0.991=7.69^o

The following vectors have been given: \vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}

The angle between these two vectors can be found by:

cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\&#10;||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}

\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54

||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}

cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o

7 0
3 years ago
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